Can you find the value of this limit in a better way?

Question

We have the sequence of real numbers $(x_{n})_{n\geq1}$ defined by $x_{1}>0$ and $$x_{n+1}=\frac{1}{n}\sum_{k=1}^n\sqrt{x_{k}^2+x_{k}}$$ for every $n\geq1$. Show that: \begin{align} \text{a)}\quad&\lim_{n\to\infty}\frac{x_{n}}{\ln n}=\frac{1}{2}\\ \text{b)}\quad&\lim_{n\to\infty}\left(\frac{x_{n}}{\ln n}-\frac{1}{2}\right)\frac{\ln n}{\ln\ln n}=-\frac{1}{4} \end{align} Ok, so I showed first by induction that $x_{n}>0$ for every $n\geq1$. I rewrote the recursion formula of the sequence: $x_{n+1} = \frac{n-1}{n}x_{n}+\frac{1}{n}\sqrt{x_{n}^2+x_{n}}$ . I showed then the sequence is inscreasing. I supposed that the sequence is upper bounded so it will converge. If the sequence converge then it is a fundamental sequence. So it exists an N natural number such that for every $n>N$ and every $\epsilon>0$ we have $\mid l - x_{n}\mid<\epsilon$, where $l$ is the limit (real number) of the sequence. By giving the value $\frac{l}{8l+2}$ for $\epsilon$ the inequality fails so the sequence is not a fundamental one, so it will not convegre wich is a contradiciton. So the sequence is not upper bounded, so the limit of $x_{n}=\infty$. For a) we can use the Stolz-Cesaro theorem and rewrite $x_{n+1}$ in terms of $x_{n}$ by the "new" recursion formula. If we use conjugation on the square root, then factor $x_{n}$, simplify the argument of the limit by $x_{n}$ and use the fact that the limit of $x_{n}=\infty$, we will get the answer $\frac{1}{2}$. b) is the main problem. Why? Because I could not find a "friendly" solution. I used twice Stolz-Cesaro Theorem and I got pages of ugly calculations and then I used some Taylor series to calculate hard limits. Yes, I got $-\frac{1}{4}$, but with an ugly method. I wonder if someone can find for b) a better solution. I will leave some photos with my ugly solution on b).

Answer 1

The initial steps are done in the right direction: we rewrite the recurrence as $$n(x_{n+1}-x_n)=\sqrt{x_n^2+x_n}-x_n,$$ observe that $x_n>0$ for each $n>0$, hence $x_{n+1}>x_n$, thus $x_n\to\infty$ as $n\to\infty$ (otherwise $x_n\to x$ for some $x>0$, which would imply $n(x_{n+1}-x_n)\to\sqrt{x^2+x}-x>0$, hence $x_{n+1}-x_n>c/n$ for some $c>0$ and sufficiently large $n$, contradiction). This way we get $$\lim_{n\to\infty}\frac{x_{n+1}-x_n}{\ln(n+1)-\ln n}=\lim_{n\to\infty}n(x_{n+1}-x_n)=\lim_{n\to\infty}\big(\sqrt{x_n^2+x_n}-x_n\big)=\frac12,$$ i.e. $x_n/\ln n\to1/2$ by Stolz–Cesàro. Put $x_n=(\ln n)/2+y_n$ with $y_n=o(\ln n)$; then \begin{gather} \frac{n}{2}\ln\left(1+\frac1n\right)+n(y_{n+1}-y_n)=x_n\left(\sqrt{1+\frac1{x_n}}-1\right);\\ n(y_{n+1}-y_n)+\frac12+O\left(\frac1n\right)=\frac12-\frac1{8x_n}+O\left(\frac1{x_n^2}\right);\\ y_{n+1}-y_n=-\frac{1+o(1)}{4n\ln n}, \end{gather} and we get $y_n/\ln\ln n\to-1/4$ as $n\to\infty$, again by Stolz–Cesàro, since $$\lim_{n\to\infty}(\ln\ln(n+1)-\ln\ln n)n\ln n=1.$$